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skelet666 [1.2K]
3 years ago
10

When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is referred to as

the?
Physics
1 answer:
Dafna11 [192]3 years ago
3 0
When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is called "load triangle". Because of a motorcycle's size and weight<span> and the fact that it has only two wheels, how to carry extra load is very important. One has to make sure that they are keeping the weight low and close to the middle of the motorcycle and keep the load evenly from side to side. Heavier items should be in the "load triangle".</span><span> </span>
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For a projectile launched from the vertically from the surface the earth with initial velocity v0, the velocity of the projectil
UNO [17]

Answer:

(dr/dt) = √(v₀² + k²/r - k²/R)

Explanation:

v² = v₀² + k²[(1/r) - (1/R)]

v = velocity of Body launched from the surface of the earth

v = initial velocity of body

k = a constant

r = distance from the centre of the earth

R = radius of the earth

v² = v₀² + k²/r - k²/R

v = √(v₀² + k²/r - k²/R)

But v = dr/dt

(dr/dt) = √(v₀² + k²/r - k²/R)

(dr/√(v₀² + k²/r - k²/R)) = dt

To go one step beyond and integrate the differential equation

∫ (dr/√(v₀² + k²/r - k²/R)) = ∫ dt

Integrating the left hand side from 0 to r and the right hand side from 0 to t

Note, v₀, k and R are all constants

(- 4r²/k²)[√(v₀² + k²/r - k²/R)] = t

√(v₀² + k²/r - k²/R) = (- k² t/4r²)

(v₀² + k²/r - k²/R) = (- k² t/4r²)²

(v₀² + k²/r - k²/R) = (k⁴t²/16r⁴)

(k⁴t²/16r⁴) + (k²/r) = [v₀² - (k²/R)]

k² [(k²t²/16r⁴) + (1/r)] = [v₀² - (k²/R)]

[(k²t²/16r⁴) + (1/r)] = [(v₀²/k²) - (1/R)]

3 0
3 years ago
The magnetic field lines due to a straight, current-carrying wire are __________. The magnetic field lines due to a straight, cu
Rama09 [41]

The magnetic field lines due to a straight, current-carrying wire are circular.

<u>Explanation:</u>

The concepts of Electromagnetism brought a new revolution to the science world. The idea is the source of many new modes of power and machines that reduces the manual work. Motors are the best example of machines that run on the concepts of electromagnetism. So the concept is that a current-carrying conductor induces a magnetic field in its nearby premise.

This magnetic field can perceive by the magnetic line of forces. Now, if we pour some iron dust around a current-carrying conductor, we'll see a concentric circular pattern around the straight wire whose centre will be at the conductor axis. The pattern of these magnetic lines of force may deflect with the variation of current in the wire but remain in the circular format.

4 0
3 years ago
Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the
balandron [24]

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

a)

  • Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .
  • This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).
  • Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.
  • According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:

       a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)

  • Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:

       v_{f} ^{2}  -v_{o} ^{2} = 2* a* \Delta x (2)

  • Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:

       \Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)

b)

  • We can find the time needed to come to an stop, applying the definition of acceleration, as follows:

       v_{f} = v_{o} + a*\Delta t (4)

  • Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
  • Replacing a and v₀ as we did in (3), we can solve for Δt as follows:

       \Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s   (5)

7 0
3 years ago
An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s
gizmo_the_mogwai [7]

Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

Q = ms\Delta T + mL

here we have

Q = 40(950)(680 - 32) + 40(450 \times 10^3)

Q = 24624 kJ + 18000 kJ

Q = 42624 kJ

Since this is the amount of aluminium per hour

so power required to melt is given by

P = \frac{Q}{t}

P = \frac{42624}{3600} kW

P = 11.84 kW

Since the efficiency is 85% so actual power required will be

P = \frac{11.84}{0.85} = 13.93 kW

Part B)

Total energy consumed by the furnace for 30 hours

Energy = power \times time

Energy = 13.93 kW\times 30 h

Energy = 417.9 kWh

now the total cost of energy consumption is given as

R = P \times 20 \frac{Cents}{kWh}

R = 417.9 kWh\times  20 \frac{cents}{kWh}

R = 8357.6 Cents

3 0
3 years ago
A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget
mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, m_{1} = 4 kg

Speed, v_{1} = 5 m/s

m_{2} = 1 kg

v_{2} = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2

By plugging in the values we get,

KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)

KE = 40J + KE(lost)

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0
3 years ago
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